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Homotopy theory, homological algebra, algebraic treatments of manifolds.
6
votes
free action on contractible spaces
The example I gave in comments has $E/G_1$ and $E/G_2$ compact (but not manifolds), at least if you use compact "blobs".
More explicitly, let $E$ be the subset of $\mathbb{R}^2$ given by the union of …
12
votes
Is every paracompact, Hausdorff, locally contractible space homotopy equivalent to a CW comp...
I think Milnor answers your question in the paper where he proves the theorem you refer to ("On spaces having the homotopy type of a CW-complex", Trans. AMS, 90 (1959), 272-280):
Whitehead had observ …
2
votes
Technical but elementary homotopy question
Can't you just take $X=[0,1]$ and $K=\{0,1\}$, with
$$f(x,t)=\begin{cases}\frac{x}{1-t}&\mbox{if $t\leq 1-x$}\\
1&\mbox{if $t\geq 1-x$},\end{cases}$$
so that $f(x,t)\to 1$ as $t\to 1$ unless $x=0$, in …
15
votes
Accepted
Which spaces have the (weak) homotopy type of compact Hausdorff spaces?
Expanding on my comment, if there are measurable cardinals then it follows from the results of
A. Przeździecki, Measurable cardinals and fundamental groups of compact spaces.
Fund. Math. 192 (2006), …
6
votes
If an abelian category $\mathcal{A}$ has enough injectives then so does $\mathrm{Ch}^{\geq 0...
If $X^{\bullet}$ is an object of $\mathrm{Ch}^{\geq0}(\mathcal{A})$ and for each $i\geq0$ $X^i\to I^i$ is an embedding into an injective, then $X^{\bullet}$ embeds in
$$I^{\bullet}:=\dots\to 0\to I^0\ …
10
votes
Accepted
Group cohomology with coefficients in a permutation module
I'm not sure I understand your intuition, but the statement is true. It's a special case of Shapiro's Lemma: if $H$ is a subgroup of $G$, $A$ is a $\mathbb{Z}H$-module, and $A^G$ is the coinduced modu …
6
votes
A sort of "group-ring" construction on coefficient systems in group homology (+ special case...
$R[M]$ is just a permutation module, so, by Shapiro's Lemma, $H_{\ast}(G;R[M])$ is the direct sum, over a set of representatives $m$ of orbits of $G$ acting on $M$, of the homology $H_{\ast}(G_m;R)$ w …
4
votes
Accepted
When are automorphisms in categories homotopically trivial?
How about adding another object $W$ to your example with an automorphism $\delta:W\to W$ with $\delta^{-1}=\delta$ and arrows $\theta,\phi:X\to W$ with $\delta\circ\theta=\phi=\theta\circ\gamma$?
The …
5
votes
Free and cellular G-action implies free G-complex?
If $G$ acts freely on a CW-complex, permuting the cells, then the stabilizer of a cell must be finite (and therefore trivial, as pointed out in the question).
This can be shown by induction on the di …
13
votes
Accepted
Z/p action on finite contractible complex
If a finite $p$-group $P$ acts on a finite-dimensional $CW$-complex $X$ which is acyclic mod $p$, then the fixed point set $X^P$ is also acyclic mod $p$. This is a special case of "Smith theory" (see …
20
votes
Accepted
Is the homotopy category of an abelian model category abelian?
No. The projective model structure on chain complexes of modules over a ring is an abelian model category, and the homotopy category is the derived category, which is never abelian unless the ring is …
6
votes
Unbounded acyclic resolutions
I'm afraid this is not very close to the case that you say you're most interested in ... maybe you want $F$ to preserve products?
But let $A=k[x]/(x^2)$, and let $\mathscr{A}$ be the category $\operat …
4
votes
Homological vs. cohomological dimension of a group/space
As noted in the question, "Question 3" reduces to asking whether, if $H_i(G,M)=0$ for all $i>0$ and all $\mathbb{Z}G$-modules $M$, then $G$ must be trivial.
But this is true, since if $G$ is non-triv …
30
votes
If $A$, $B$ are abelian groups such that $\mathrm{Hom}(A, G) \cong \mathrm{Hom}(B, G)$ for a...
I just stumbled across the answer to this in Fuchs' 2015 book on Abelian Groups.
The papers
Hill, Paul, Two problems of Fuchs concerning tor and hom, J. Algebra 19, 379-383 (1971). ZBL0228.20027.
and …